I'm working on a few scripts where I need to check for a few environment variables and list all the ones missing. I'm seeing a lot of posts where it would check for one and exit.
if [ -z "$BLAH" ]; then
echo "Missing BLAH"
exit 1
fi
However, I would like to print all the missing ones and then exit if anything is not set. I'm doing something like this right now, Is there any more elegant way to do this?
function check_env_vars {
status=0
for name in $*; do
value="${!name}"
echo "$value"
if [[ -z "$value" ]]; then
echo "$name environment variable must not be empty"
status=1
fi
done
return $status
}
if [[ check_env_vars "BLAH" "BLAH1" "BLAH2" -ne 0 ]]; then
exit 1
fi
Appreciate any thoughts or ideas.
if already checks the exit status; that's what it does with [[ ... ]] in the first place.
if ! check_env_vars "BLAH" "BLAH1" "BLAH2"; then
exit 1
fi
That said, bash already has syntax for verifying that a variable is set and non-null:
check_vars () {
for name; do
: ${!name:?$name must not be empty}
done
}
Related
What would be the best way to check the exit status in an if statement in order to echo a specific output?
I'm thinking of it being:
if [ $? -eq 1 ]
then
echo "blah blah blah"
fi
The issue I am also having is that the exit statement is before the if statement simply because it has to have that exit code. Also, I know I'm doing something wrong since the exit would obviously exit the program.
Every command that runs has an exit status.
That check is looking at the exit status of the command that finished most recently before that line runs.
If you want your script to exit when that test returns true (the previous command failed) then you put exit 1 (or whatever) inside that if block after the echo.
That being said, if you are running the command and are wanting to test its output, using the following is often more straightforward.
if some_command; then
echo command returned true
else
echo command returned some error
fi
Or to turn that around use ! for negation
if ! some_command; then
echo command returned some error
else
echo command returned true
fi
Note though that neither of those cares what the error code is. If you know you only care about a specific error code then you need to check $? manually.
Note that exit codes != 0 are used to report errors. So, it's better to do:
retVal=$?
if [ $retVal -ne 0 ]; then
echo "Error"
fi
exit $retVal
instead of
# will fail for error codes == 1
retVal=$?
if [ $retVal -eq 1 ]; then
echo "Error"
fi
exit $retVal
An alternative to an explicit if statement
Minimally:
test $? -eq 0 || echo "something bad happened"
Complete:
EXITCODE=$?
test $EXITCODE -eq 0 && echo "something good happened" || echo "something bad happened";
exit $EXITCODE
$? is a parameter like any other. You can save its value to use before ultimately calling exit.
exit_status=$?
if [ $exit_status -eq 1 ]; then
echo "blah blah blah"
fi
exit $exit_status
For the record, if the script is run with set -e (or #!/bin/bash -e) and you therefore cannot check $? directly (since the script would terminate on any return code other than zero), but want to handle a specific code, #gboffis comment is great:
/some/command || error_code=$?
if [ "${error_code}" -eq 2 ]; then
...
Just to add to the helpful and detailed answer:
If you have to check the exit code explicitly, it is better to use the arithmetic operator, (( ... )), this way:
run_some_command
(($? != 0)) && { printf '%s\n' "Command exited with non-zero"; exit 1; }
Or, use a case statement:
run_some_command; ec=$? # grab the exit code into a variable so that it can
# be reused later, without the fear of being overwritten
case $ec in
0) ;;
1) printf '%s\n' "Command exited with non-zero"; exit 1;;
*) do_something_else;;
esac
Related answer about error handling in Bash:
Raise error in a Bash script
If you are writing a function – which is always preferred – you can propagate the error like this:
function()
{
if <command>; then
echo worked
else
return
fi
}
Now, the caller can do things like function && next as expected! This is useful if you have a lot of things to do in the if block, etc. (otherwise there are one-liners for this). It can easily be tested using the false command.
Using Z shell (zsh) you can simply use:
if [[ $(false)? -eq 1 ]]; then echo "yes" ;fi
When using Bash and set -e is on, you can use:
false || exit_code=$?
if [[ ${exit_code} -ne 0 ]]; then echo ${exit_code}; fi
This might only be useful in a limited set of use-cases, I use this specifically when I need to capture the output from a command and write it to a log file if the exit code reports that something went wrong.
RESULT=$(my_command_that_might_fail)
if (exit $?)
then
echo "everything went fine."
else
echo "ERROR: $RESULT" >> my_logfile.txt
fi
you can just add this if statement:
if [ $? -ne 0 ];
then
echo 'The previous command was not executed successfully';
fi
Below test scripts below work for
simple bash test commands
multiple test commands
bash test commands with pipe included:
if [[ $(echo -en "abc\n def" |grep -e "^abc") && ! $(echo -en "abc\n def" |grep -e "^def") ]] ; then
echo "pipe true"
else
echo "pipe false"
fi
if [[ $(echo -en "abc\n def" |grep -e "^abc") && $(echo -en "abc\n def" |grep -e "^def") ]] ; then
echo "pipe true"
else
echo "pipe false"
fi
The output is:
pipe true
pipe false
I want to check, if multiple variable are set or not, if set then only execute the script code, otherwise exit.
something like:
if [ ! $DB=="" && $HOST=="" && $DATE=="" ]; then
echo "you did not set any variable"
exit 1;
else
echo "You are good to go"
fi
You can use -z to test whether a variable is unset or empty:
if [[ -z $DB || -z $HOST || -z $DATE ]]; then
echo 'one or more variables are undefined'
exit 1
fi
echo "You are good to go"
As you have used the bash tag, I've used an extended test [[, which means that I don't need to use quotes around my variables. I'm assuming that you need all three variables to be defined in order to continue. The exit in the if branch means that the else is superfluous.
The standard way to do it in any POSIX-compliant shell would be like this:
if [ -z "$DB" ] || [ -z "$HOST" ] || [ -z "$DATE" ]; then
echo 'one or more variables are undefined'
exit 1
fi
The important differences here are that each variable check goes inside a separate test and that double quotes are used around each parameter expansion.
If you are ok with writing a function for this purpose, it can be pretty convenient.
This solution uses the ${!VAR_NAME} syntax to check whether the variable is empty and has the added benefit of telling you which variable names are empty.
check_vars()
{
var_names=("$#")
for var_name in "${var_names[#]}"; do
[ -z "${!var_name}" ] && echo "$var_name is unset." && var_unset=true
done
[ -n "$var_unset" ] && exit 1
return 0
}
# Usage for this case
check_vars DB HOST DATE
echo "You are good to go"
I wound up using variable-variables to loop through an easily managed HEREDOC list of variable names:
# Ensure non-empty values.
# Loop through HEREDOC, test variable-variable isn't blank.
while read var; do
[ -z "${!var}" ] && { echo "$var is empty or not set. Exiting.."; exit 1; }
done << EOF
KUBE_NAMESPACE
DOCKER_REGISTRY
DOCKER_DEPLOY_USER
DOCKER_DEPLOY_PASSWORD
DOCKER_DEPLOY_EMAIL
EOF
You can check it also by put the variables name in a file
DB=myDB
HOST=myDB
DATE=myDATE
then test them if currently empty or unset
#!/bin/bash
while read -r line; do
var=`echo $line | cut -d '=' -f1`
test=$(echo $var)
if [ -z "$(test)" ]; then
echo 'one or more variables are undefined'
exit 1
fi
done <var.txt
echo "You are good to go"
Nice solution from #joe.still !
improvement is to exit after checking all variables
i=0
while read var; do
[ -z "${!var}" ] && { echo "$var is empty or not set. Exiting.."; let i=i+1; }
done << EOF
KUBE_NAMESPACE
DOCKER_REGISTRY
DOCKER_DEPLOY_USER
DOCKER_DEPLOY_PASSWORD
DOCKER_DEPLOY_EMAIL
EOF
if [ $i -gt 0 ]; then
echo $i
echo "exiting"
exit 1
fi
Good Day Everyone.
I've personally used this method in my bash scripts. Verified works on bash 4.4 and later in Ubuntu, openSUSE, and ClearLinux.
Can RHEL|CentOS|Alma and Arch Based users let me know it it works fine for you?
( [ "$VAR1""$VAR2""$VAR3""$VAR4""$VAR5" ] && echo -e " Warning: StackIsNotClear" ) || { echo -e " GoodNews: StackIsClear"; }
I am trying to perform this:
i have a test file which md5sum of files located on sftp.
variables should contain an md5sum (string), if the variable is empty it means there is no file on the sftp server.
i am trying this code but it does not work..
if [ -z $I_IDOCMD5 ] || [ -z $I_LEGALMD5 ] || [ -z $I_ZIPMD5 ]
then
echo "ERROR: At least one file not present of checksum missing no files will be deleted" >>$IN_LOG
ERRORS=$ERRORS+2
else
if [[ $I_IDOCMD5 == $($DIGEST -a md5 $SAPFOLDER/inward/idoc/$I_IDOC) ]]
then
echo "rm IDOC/$I_IDOC" >/SAP/commands_sftp.in
else
echo "problem with checksum"
ERRORS=$ERRORS+2
fi
if [[ $I_LEGALMD5 == $($DIGEST -a md5 $SAPFOLDER/inward/legal/$I_LEGAL) ]]
then
echo "rm LEGAL/$I_LEGAL" >>/SAP/commands_sftp.in
else
echo "problem with checksum"
ERRORS=$ERRORS+2
fi
if [[ $I_ZIPMD5 == $($DIGEST -a md5 $SAPFOLDER/inward/zip/$I_ZIP) ]]
then
echo "rm ZIP/$I_ZIP" >>/SAP/commands_sftp.in
else
echo "problem with checksum"
ERRORS=$ERRORS+2
fi
The answer I prefer is following
[[ -z "$1" ]] && { echo "Parameter 1 is empty" ; exit 1; }
Note, don't forget the ; into the {} after each instruction
One way to check if a variable is empty is:
if [ "$var" = "" ]; then
# $var is empty
fi
Another, shorter alternative is this:
[ "$var" ] || # var is empty
In bash you can use set -u which causes bash to exit on failed parameter expansion.
From bash man (section about set builtin):
-u
Treat unset variables and parameters other than the special parameters "#" and "*" as an error when performing parameter
expansion. If expansion is attempted on an unset variable or
parameter, the shell prints an error message, and, if not interactive,
exits with a non-zero status.
For more information I recommend this article:
http://redsymbol.net/articles/unofficial-bash-strict-mode/
You can use a short form:
FNAME="$I_IDOCMD5"
: ${FNAME:="$I_LEGALMD5"}
: ${FNAME:="$I_ZIPMD5"}
: ${FNAME:?"Usage: $0 filename"}
In this case the script will exit if neither of the I_... variables is declared, printing an error message prepended with the shell script line that triggered the message.
See more on this in abs-guide (search for «Example 10-7»).
First test only this (just to narrow it down):
if [ -z "$I_IDOCMD5" ] || [ -z "$I_LEGALMD5" ] || [ -z "$I_ZIPMD5" ]
then
echo "one is missing"
else
echo "everything OK"
fi
echo "\"$I_IDOCMD5\""
echo "\"$I_LEGALMD5\""
echo "\"$I_ZIPMD5\""
"if the variable is empty it means there is no file on the sftp server"
If there is no file on the sftp server, is the variable then really empty ?
No hidden spaces or anything like that ? or the number zero (which counts as non-empty) ?
I am looking for suggestions for a solution and on best approaches to handle figuring out if multiple IFs are null.
I have:
if [ -n "$sfcompname" ]; then
echo $sfcompname
fi
if [ -n "$sfcompip" ]; then
echo $sfcompip
fi
if [ -n "$lacompname" ]; then
echo $lacompname
fi
if [ -n "$lacompip" ]; then
echo $lacompip
fi
.. I'm sure that can be done better, but my main problem at the moment is trying to then say:
if (all those IFs) = null
echo "Please check the name you entered and try again"
Somewhat silly, but should work
if ! [[ ${sfcompname}${sfcompip}${lacompname}${lacompip} ]]
then
echo "Please check the name you entered and try again"
fi
You can use another variable for this which you initialise to a value and then change if any of the if statements fire. Then at the end, if it hasn't changed, then you know that none of them fired. Something like this will suffice:
fired=0
if [ -n "$sfcompname" ]; then
echo $sfcompname
fired=1
fi
if [ -n "$sfcompip" ]; then
echo $sfcompip
fired=1
fi
if [ -n "$lacompname" ]; then
echo $lacompname
fired=1
fi
if [ -n "$lacompip" ]; then
echo $lacompip
fired=1
fi
if [[ ${fired} -eq 0 ]] ; then
echo 'None were fired'
fi
Another possibility is to use the variable check short-cut:
name="$sfcompname$sfcompip$lacompname$lacompip"
${name:?"Please check the name you entered and try again"}
This will exit the program if none of the variables are set. The message is optional, it overrides the standard "parameter null or not set".
I am trying to write a bash script that takes in an option.
Lets call these options A and B.
In the script A and B may or may not be defined as variables.
I want to be able to check if the variable is defined or not.
I have tried the following but it doesn't work.
if [ ! -n $1 ]; then
echo "Error"
fi
Thanks
The "correct" way to test whether a variable is set is to use the + expansion option. You'll see this a lot in configure scripts:
if test -s "${foo+set}"
where ${foo+set} expands to "set" if it is set or "" if it's not. This allows for the variable to be set but empty, if you need it. ${foo:+set} additionally requires $foo to not be empty.
(That $(eval echo $a) thing has problems: it's slow, and it's vulnerable to code injection (!).)
Oh, and if you just want to throw an error if something required isn't set, you can just refer to the variable as ${foo:?} (leave off the : if set but empty is permissible), or for a custom error message ${foo:?Please specify a foo.}.
You did not define how these options should be passed in, but I think:
if [ -z "$1" ]; then
echo "Error"
exit 1
fi
is what you are looking for.
However, if some of these options are, err, optional, then you might want something like:
#!/bin/bash
USAGE="$0: [-a] [--alpha] [-b type] [--beta file] [-g|--gamma] args..."
ARGS=`POSIXLY_CORRECT=1 getopt -n "$0" -s bash -o ab:g -l alpha,beta:,gamma -- "$#"`
if [ $? -ne 0 ]
then
echo "$USAGE" >&2
exit 1
fi
eval set -- "$ARGS"
unset ARGS
while true
do
case "$1" in
-a) echo "Option a"; shift;;
--alpha) echo "Option alpha"; shift;;
-b) echo "Option b, arg '$2'"; shift 2;;
--beta) echo "Option beta, arg '$2'"; shift 2;;
-g|--gamma) echo "Option g or gamma"; shift;;
--) shift ; break ;;
*) echo "Internal error!" ; exit 1 ;;
esac
done
echo Remaining args
for arg in "$#"
do
echo '--> '"\`$arg'"
done
exit 0
Don't do it that way, try this:
if [[ -z $1 ]]; then
echo "Error"
fi
The error in your version is actually the lack of quoting.
Should be:
if [ ! -n "$1" ]; then
echo "Error"
fi
But you don't need the negation, use -z instead.
If you work on Bash, then use double brackets [[ ]] too.
from the man bash page:
-z string
True if the length of string is zero.
-n string
True if the length of string is non-zero.
Also, if you use bash v4 or greater (bash --version) there's -v
-v varname
True if the shell variable varname is set (has been assigned a value).
The trick is "$1", i.e.
root#root:~# cat auto.sh
Usage () {
echo "error"
}
if [ ! -n $1 ];then
Usage
exit 1
fi
root#root:~# bash auto.sh
root#root:~# cat auto2.sh
Usage () {
echo "error"
}
if [ ! -n "$1" ];then
Usage
exit 1
fi
root#root:~# bash auto2.sh
error